What are the solutions to the equation 4x^2-8x+4=0 ?

The solutions to the equation 4x^2-8x+4=0 are x=1

Find the zeros of 4x^2-8x+4=0

The zeros of 4x^2-8x+4=0 are x=1

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4 x^{2} - 8 x + 4 = 0

x = 1

Solution steps

4 x^{2} - 8 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

(- 8)^{2} - 4 \cdot 4 \cdot 4 = 0

x_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 4}

x = \frac{- ( - 8 )}{2 \cdot 4}

\frac{- ( - 8 )}{2 \cdot 4} = 1

x = 1

The solution to the quadratic equation is:

x = 1

\frac{3}{x ^{2} - 1} + 1 > \frac{2}{x - 1}

6 - 3 + 4 x + 1 = 5 x + 7

- x + 8 > 6

3 x^{2} + 8 x \leq 0

4 x^{2} - 12 x + 25 = 0

What are the solutions to the equation 4x^2-8x+4=0 ?
The solutions to the equation 4x^2-8x+4=0 are x=1
Find the zeros of 4x^2-8x+4=0
The zeros of 4x^2-8x+4=0 are x=1